MathDoofus wrote:So how does induction prove anything beyond the relationship between the if statement in step 1 and the rewritten equation in step 4?

It doesn't. That's why you need the base step.

The induction step says that, if you had a rung to stand on, you could get to the next one. The base step says - here, you can stand on this rung. Neither of these parts does anything by themselves, but together they let you climb the ladder.

No, even in theory, you cannot build a rocket more massive than the visible universe.

MathDoofus wrote:So how does induction prove anything beyond the relationship between the if statement in step 1 and the rewritten equation in step 4?

It doesn't. That's why you need the base step.

The induction step says that, if you had a rung to stand on, you could get to the next one. The base step says - here, you can stand on this rung. Neither of these parts does anything by themselves, but together they let you climb the ladder.

MathDoofus wrote:So how does induction prove anything beyond the relationship between the if statement in step 1 and the rewritten equation in step 4?

It doesn't. That's why you need the base step.

The induction step says that, if you had a rung to stand on, you could get to the next one. The base step says - here, you can stand on this rung. Neither of these parts does anything by themselves, but together they let you climb the ladder.

But why do we get to assume both the k case and the (k + 1) case?

Where do we assume the (k + 1) case?

gmalivuk wrote:

King Author wrote:If space (rather, distance) is an illusion, it'd be possible for one meta-me to experience both body's sensory inputs.

Yes. And if wishes were horses, wishing wells would fill up very quickly with drowned horses.

MathDoofus wrote:So how does induction prove anything beyond the relationship between the if statement in step 1 and the rewritten equation in step 4?

It doesn't. That's why you need the base step.

The induction step says that, if you had a rung to stand on, you could get to the next one. The base step says - here, you can stand on this rung. Neither of these parts does anything by themselves, but together they let you climb the ladder.

Induction doesn't depend on the countability of natural numbers, but on their well-ordering. In a well-ordered set, every nonempty subset has a smallest element.

In particular, this means that if there are any counterexamples to what you want to prove, there must be a smallest counterexample.

The smallest counterexample isn't 1 (or 2 or 3 or 4, in this case), because you proved the statement for 1 with the base case. So the smallest counterexample, if it exists, must be k+1 for some larger integer k. Because we're thinking about the *smallest* counterexample, we know the statement must be true for k.

But then we prove that IF it's true for k, THEN it must also be true for k+1. So k+1 can't be a counterexample. And because our proof doesn't care what k is, that means NO integer can be a counterexample.---Honestly, though, it seems like your misunderstandings are too basic for you to try to jump right to trying to understand induction.

Make sure first that you understand how conditional statements work logically, and that you understand what sorts of operations you can do in proofs that preserve equality (that is, make sure you understand why it's okay to add k+1 to both sides of an equation).

As it is, you seem to keep going back and forth between high level questions about how assuming it's true for k is non-circular for the induction step, and low-level questions like how we can get from (k+1)+1 to k+2.

Unless stated otherwise, I do not care whether a statement, by itself, constitutes a persuasive political argument. I care whether it's true.---If this post has math that doesn't work for you, use TeX the World for Firefox or Chrome

MathDoofus wrote:Can you explain how the k and (k + 1) case fit with modus ponens and prove whatever it is we're trying to show for all instances of the base case and above?

I'm going to prove the 4-case, using the base step and the induction step. The induction step holds for all k, so I can plug in any specific value of k and it will still hold.

The 1-case holds by the base step.

Taking k=1, the induction steps says that if the 1-case holds, then the 2-case holds. So by modus ponens, the 2-case holds.

Now we use the induction step again with k=2: if the 2-case holds, then the 3-case holds. So by modus ponens, the 3-case holds.

Now we use the induction step again with k=3: if the 3-case holds, then the 4-case holds. So by modus ponens, the 4-case holds.

See how I could just add more steps of this to prove the 5-case, or the 100-case?

How do the base case and the k case demonstrate modus ponens? I'm not sure what the k case is supposed to demonstrate other than k can be substituted for n and various algebraic stuff can be performed after the substitution.

Honestly, though, it seems like your misunderstandings are too basic for you to try to jump right to trying to understand induction.

Make sure first that you understand how conditional statements work logically, and that you understand what sorts of operations you can do in proofs that preserve equality (that is, make sure you understand why it's okay to add k+1 to both sides of an equation).

As it is, you seem to keep going back and forth between high level questions about how assuming it's true for k is non-circular for the induction step, and low-level questions like how we can get from (k+1)+1 to k+2.

You're probably right. I have a learning disability relating to math and the manipulation/understanding of abstract symbols because of a large gulf between performance IQ and verbal IQ. I am trying to re-teach this stuff to myself to prove that it's graspable. My algebra is rusty (as you've seen) and proof by induction still strikes me as unsatisfactory in some way that I can't identify precisely.

I think it's more fundamental than some rusty algebra, though maybe you understand the concept and just have trouble parsing what's happening in the equations.

If two things are equal, then we can do the same thing to both of them and they remain equal. One thing we can do is add k+1 to both of them.

In the induction step, we just need k to be *some* number for which the equation is true. We know there are some numbers like this, because that's what we showed in the base case. So you write the equation for that number, and then the induction step shows that the equation must also be true for the next number.

By itself the induction step is useless. If I tell you, "If it's raining then I have an umbrella," that's not enough information for you to conclude that I have an umbrella today. But if I also tell you that it's raining today, now you can conclude that I have an umbrella.

The induction step tells you, "If this equation is true for one number, then it is also true for the next number." The base case tells you, "This equation is true for at least one number." So now we have the whole chain. It's true for 1, so it must be true for 2, so it must be true for 3, so it must be true for 4. The natural numbers are well-ordered, which means all of them are somewhere in this sequence.

Unless stated otherwise, I do not care whether a statement, by itself, constitutes a persuasive political argument. I care whether it's true.---If this post has math that doesn't work for you, use TeX the World for Firefox or Chrome

gmalivuk wrote:I think it's more fundamental than some rusty algebra, though maybe you understand the concept and just have trouble parsing what's happening in the equations.

If two things are equal, then we can do the same thing to both of them and they remain equal. One thing we can do is add k+1 to both of them.

In the induction step, we just need k to be *some* number for which the equation is true. We know there are some numbers like this, because that's what we showed in the base case. So you write the equation for that number, and then the induction step shows that the equation must also be true for the next number.

By itself the induction step is useless. If I tell you, "If it's raining then I have an umbrella," that's not enough information for you to conclude that I have an umbrella today. But if I also tell you that it's raining today, now you can conclude that I have an umbrella.

The induction step tells you, "If this equation is true for one number, then it is also true for the next number." The base case tells you, "This equation is true for at least one number." So now we have the whole chain. It's true for 1, so it must be true for 2, so it must be true for 3, so it must be true for 4. The natural numbers are well-ordered, which means all of them are somewhere in this sequence.

And to build on that, because I used 1 in the base case, I've found the smallest example such that (1) there is no smaller counterexample and (2) it must be true of everything greater than 1 because of the axiom that the natural numbers are well-ordered. Is that right? And the "k" case need not be the "k" case, I could choose any placeholder ("n," say). Is that right?

I am going to work back through the original example and post my work.

Also, is there another entry-level example that I could work on? I did some poking around online but many of the examples look too complex (whether because of exponent notation, summation notation, or other symbols that are really unfamiliar at this point in my life). Thanks to all for their help.

Last edited by MathDoofus on Tue May 02, 2017 2:15 pm UTC, edited 1 time in total.

Meteoric wrote:Honestly, the more we work on this problem, the more I think this is absolutely terrible as a first example of induction

Yup, too bad it is the example for induction. (I think it does work the other way around: induction is a great way to rigorously show that 1+2+...+n=n(n+1)/2)

This problem may sound a bit trivial, but eh, it's something (can you recognize the last math topic I had was markov chains?):Say I want to go on holiday to a town in Italy some time in the future. For some reason the weather for the next day there is completely predictable every day:

Of course, it's a holiday, so I only want to go there if it's sunny. The weather report of 2 days ago said it was raining.Will I ever go to that town?

Note that induction works with "if something is true on start day" and "if something is true on some day then something must be true on the next day", so you have to rewrite the negations in the problem statement.If that's troubling, here's the question in a more direct way:

Spoiler:

The weather report said it was not sunny. Will it always be not sunny?

[edit]rephrased a bit

Last edited by Flumble on Tue May 02, 2017 5:18 pm UTC, edited 3 times in total.

As promised, here's what I've got so far in using pen and paper and working back through the original problem in light of everyone's help.

I want to show the following: 1 + 2 + 3 + ... + n = n(n + 1)/2

First, I'll perform a check using 1 as a base case, to create the first "rung" on the ladder that I'm hoping will result.

1 = 1(1 + 1)/21 = 1(2)/21 = 2/21 = 1

The base case checks out.

Second, I'll assume that it's true for the k case. In other words, I'll use k as a placeholder for any natural number, and see what happens.

1 + 2 + 3 + ... + k = k(k + 1)/2

Again, the above is an assumption. I'm using it so that I can show that if something is true of the k case, then it's also true of the (k + 1) case.

Third, I'll add (k + 1) to both sides of the above-referenced equation. The reason that I can be absolutely sure that the (k + 1) version is equal to the version containing only k is because I'm adding an identical element (k + 1) to both sides of an equation.

Also, to follow up on the above post, why is it that I'm adding (k + 1) to both sides? And did I add (k + 1) to both sides correctly? Or should I substitute, on the right-hand side, (k + 1) for every instance of k? I could use some help because I'm now stuck. Thank you.

That's not exactly the way I would do this, but I don't necessarily want to point you in a different direction from what you've been told. What I do is try to actually write n in the original equation with "k+1". So I get:1+2+...+k+(k+1)=(k+1)*(k+1 +1)/2I don't know that this is true yet, I'm trying to prove it.

Zohar wrote:That's not exactly the way I would do this, but I don't necessarily want to point you in a different direction from what you've been told. What I do is try to actually write n in the original equation with "k+1". So I get:1+2+...+k+(k+1)=(k+1)*(k+1 +1)/2I don't know that this is true yet, I'm trying to prove it.

MathDoofus wrote:As promised, here's what I've got so far in using pen and paper and working back through the original problem in light of everyone's help.

I want to show the following: 1 + 2 + 3 + ... + n = n(n + 1)/2

First, I'll perform a check using 1 as a base case, to create the first "rung" on the ladder that I'm hoping will result.

1 = 1(1 + 1)/21 = 1(2)/21 = 2/21 = 1

The base case checks out.

Second, I'll assume that it's true for the k case. In other words, I'll use k as a placeholder for any natural number, and see what happens.

1 + 2 + 3 + ... + k = k(k + 1)/2

Again, the above is an assumption. I'm using it so that I can show that if something is true of the k case, then it's also true of the (k + 1) case.

Third, I'll add (k + 1) to both sides of the above-referenced equation. The reason that I can be absolutely sure that the (k + 1) version is equal to the version containing only k is because I'm adding an identical element (k + 1) to both sides of an equation.

1 + 2 + 3 + ... + k + (k + 1) = k(k + 1)/2 + (k + 1)

Do I have it all right so far?

Everything to this point is correct. Remember, k is just a placeholder for some number, so (k+1) is just a placeholder for some slightly different number. If you add the same number to both sides of an equality, it stays equal.

The reason you add (k+1) to both sides:

1 + 2 + 3 + ... + k = k(k + 1)/2 is an equation whose left-hand side is just the sum of the first k positive integers. We can turn it into the sum of the first (k+1) integers by adding the (k+1)th integer, which is just (k+1). But if we add (k+1) to the left-hand side, we have to add it to the right-hand side, too, to make sure both sides remain equal.

Why do we want to do this? Our goal is to say that, if we assume some fact about the k case is true (namely, that 1 + 2 + 3 + ... + k = k(k + 1)/2), then it's also true of the (k+1) case. We're interested in the sum of the first k integers; we are assuming 1 + 2 + 3 + ... + k = k(k + 1)/2. The (k+1) case is the sum of the first (k+1) integers, which we get to by adding (k+1) to both sides.

MathDoofus wrote:Is that correct? I must confess to being confused yet again.

Yeah I apologize for that. WibblyWobbly and I are describing slightly different ways to approach the problem, both are valid. But I think you should just stick with where you started and ignore my post.

WibblyWobbly wrote:Everything to this point is correct. Remember, k is just a placeholder for some number, so (k+1) is just a placeholder for some slightly different number. If you add the same number to both sides of an equality, it stays equal.

The reason you add (k+1) to both sides:

1 + 2 + 3 + ... + k = k(k + 1)/2 is an equation whose left-hand side is just the sum of the first k positive integers. We can turn it into the sum of the first (k+1) integers by adding the (k+1)th integer, which is just (k+1). But if we add (k+1) to the left-hand side, we have to add it to the right-hand side, too, to make sure both sides remain equal.

Why do we want to do this? Our goal is to say that, if we assume some fact about the k case is true (namely, that 1 + 2 + 3 + ... + k = k(k + 1)/2), then it's also true of the (k+1) case. We're interested in the sum of the first k integers; we are assuming 1 + 2 + 3 + ... + k = k(k + 1)/2. The (k+1) case is the sum of the first (k+1) integers, which we get to by adding (k+1) to both sides.

Great! And, to add to your post with what (I think) I've learned so far, we're really testing the k case and the (k + 1) case at the same time because we're positing that the k case is true (through the statement 1 + 2 + 3 + ... + k = k(k + 1)/2 at the second step) and saying that, if the k case is true, then the (k + 1) case must also be true. Is that right?

Now I'm stuck; I'm not exactly sure what the next step should be after adding (k + 1) to both sides of the equation. I do recall that, at some point, I should substitute (on the left-hand side) the k(k + 1)/2 stuff (i.e., the right-hand side of the second step) because we've assumed that it equals the k case. But beyond that - please help!

MathDoofus wrote:How do the base case and the k case demonstrate modus ponens? I'm not sure what the k case is supposed to demonstrate other than k can be substituted for n and various algebraic stuff can be performed after the substitution.

I think you are mixing up implications with their components here. We did not prove the k case. We proved "for all values of k, k case ⇒ k+1 case". Then we used modus ponens because that is how modus ponens works: you have A, and you have A⇒B, and you conclude B.

If it rains tomorrow, then the sidewalk will be wet. This sentence is not a weather forecast: I don't know if it will rain. It's not a prediction about the sidewalk either: I don't know whether the sidewalk will be wet or dry. It is a statement about the relationship between those things. Maybe it won't rain, I don't know - but if it does, the sidewalk will definitely be wet.

I would strongly recommend doing some non-induction practice with conditionals and proofs thereof. If you don't have that sorted out, induction won't make any sense.

No, even in theory, you cannot build a rocket more massive than the visible universe.

MathDoofus wrote:How do the base case and the k case demonstrate modus ponens? I'm not sure what the k case is supposed to demonstrate other than k can be substituted for n and various algebraic stuff can be performed after the substitution.

I think you are mixing up implications with their components here. We did not prove the k case. We proved "for all values of k, k case ⇒ k+1 case". Then we used modus ponens because that is how modus ponens works: you have A, and you have A⇒B, and you conclude B.

If it rains tomorrow, then the sidewalk will be wet. This sentence is not a weather forecast: I don't know if it will rain. It's not a prediction about the sidewalk either: I don't know whether the sidewalk will be wet or dry. It is a statement about the relationship between those things. Maybe it won't rain, I don't know - but if it does, the sidewalk will definitely be wet.

Yes. And the base case gets us the "forecast" (so to speak) because it helps show that it works in actuality, the k case shows that it works for any arbitrary number, and the (k + 1) case shows that one can always prove it for the next number in the sequence no matter which k one starts from.

Meteoric wrote:By the way, for some induction examples with less algebra to keep track of:

Prove that every natural number is either even or odd. (Hint: at the induction step, you will start by saying "assume k is either even or odd". Then, split your proof into cases.)

Prove that for every natural n, 4^n - 1 is divisible by 3.

Can you help with the fourth step above? I'd like to get the canonical example crystal-clear for myself before tackling a new problem. Thank you (and I have to confess that both of the new problems you've given me look pretty intimidating - I don't know how I would even begin to effectively notate--much less analyze--the even-or-odd designation and the "divisible by three" problem looks like it depends on an understanding of division that I don't have (or maybe even modulo or remainders, yikes!).

MathDoofus wrote:Great! And, to add to your post with what (I think) I've learned so far, we're really testing the k case and the (k + 1) case at the same time because we're positing that the k case is true (through the statement 1 + 2 + 3 + ... + k = k(k + 1)/2 at the second step) and saying that, if the k case is true, then the (k + 1) case must also be true. Is that right?

Yes, but I would not describe this as "testing the k case". We are not testing the k case - we are just examining what would happen, if the k case were true.

Now I'm stuck; I'm not exactly sure what the next step should be after adding (k + 1) to both sides of the equation. I do recall that, at some point, I should substitute (on the left-hand side) the k(k + 1)/2 stuff (i.e., the right-hand side of the second step) because we've assumed that it equals the k case. But beyond that - please help!

It seems to me that you are struggling here because you are losing track of the goal. We've had a lot of detours in this conversation, so that is very understandable. But if you ever find yourself saying "what should I do next", back up and think about what you are trying to accomplish. "Keep your eye on the ball" is surprisingly relevant advice for writing proofs.

Can you write down (a) the equation you have at this step, and (b) the equation you are trying to prove?

No, even in theory, you cannot build a rocket more massive than the visible universe.

MathDoofus wrote:Great! And, to add to your post with what (I think) I've learned so far, we're really testing the k case and the (k + 1) case at the same time because we're positing that the k case is true (through the statement 1 + 2 + 3 + ... + k = k(k + 1)/2 at the second step) and saying that, if the k case is true, then the (k + 1) case must also be true. Is that right?

Yes, but I would not describe this as "testing the k case". We are not testing the k case - we are just examining what would happen, if the k case were true.

That makes sense - the testing occurs at the base-case level, right? The base case, plus the k case (i.e., any arbitrary number), and the (k + 1) case (because you can always get to the next-greatest natural number when one starts at any k) do the work together, yes?

Now I'm stuck; I'm not exactly sure what the next step should be after adding (k + 1) to both sides of the equation. I do recall that, at some point, I should substitute (on the left-hand side) the k(k + 1)/2 stuff (i.e., the right-hand side of the second step) because we've assumed that it equals the k case. But beyond that - please help!

It seems to me that you are struggling here because you are losing track of the goal. We've had a lot of detours in this conversation, so that is very understandable. But if you ever find yourself saying "what should I do next", back up and think about what you are trying to accomplish. "Keep your eye on the ball" is surprisingly relevant advice for writing proofs.

Can you write down (a) the equation you have at this step, and (b) the equation you are trying to prove?

Yes, at the fourth step, I have the following:

1 + 2 + 3 + ... + k + (k + 1) = k(k + 1)/2 + (k + 1)

Not sure what you're asking about with respect to the "equation [I'm] trying to prove."

MathDoofus wrote:That makes sense - the testing occurs at the base-case level, right? The base case, plus the k case (i.e., any arbitrary number), and the (k + 1) case (because you can always get to the next-greatest natural number when one starts at any k) do the work together, yes?

Right, that's a much better way to phrase it.

Yes, at the fourth step, I have the following:

1 + 2 + 3 + ... + k + (k + 1) = k(k + 1)/2 + (k + 1)

Not sure what you're asking about with respect to the "equation [I'm] trying to prove."

Back up and think about it for a few minutes - I could tell you directly, but I don't think that would help as much. It's very important that you know what is being proven.

At the induction step, your goal is to prove that "k case ⇒ k+1 case". In this specific problem, what is the k+1 case?

No, even in theory, you cannot build a rocket more massive than the visible universe.

MathDoofus wrote:The (k + 1) case would be replacing all references to k with (k + 1), yes? But I'm not sure whether I do that on both the left and right sides of the equation above!

What exactly does "replacing all references" mean? You're not wrong, you just need to be more precise. The (k+1) case is a thing that you can talk about all on its own - can you write out the equation for the (k+1) case?

No, even in theory, you cannot build a rocket more massive than the visible universe.

MathDoofus wrote:The (k + 1) case would be replacing all references to k with (k + 1), yes? But I'm not sure whether I do that on both the left and right sides of the equation above!

What exactly does "replacing all references" mean? You're not wrong, you just need to be more precise. The (k+1) case is a thing that you can talk about all on its own - can you write out the equation for the (k+1) case?